# Ellipse with rectangular frame

• Jan 24th 2009, 04:49 AM
courteous
Ellipse with rectangular frame
»Write down the equation of an ellipse, that goes through $\displaystyle P(1,-2)$ and whose circumscribed* rectangle's area is minimized.«
*http://www.shrani.si/f/3e/vc/1Wz3BdGp/ellipse.gif

(Pretending not knowing that "rectangle" is in fact square.)

Set-up:
Ellipse through $\displaystyle (1,-2)$: $\displaystyle \frac{1}{a^2}+\frac{4}{b^2}=1$, rectangle's area: $\displaystyle 2a*2b=min.$
Now, what to differentiate? From the first equation I've also got $\displaystyle a^2b^2=b^2+4a^2$.
• Jan 24th 2009, 06:52 AM
skeeter
$\displaystyle a^2b^2 = b^2 + 4a^2$

$\displaystyle a^2b^2 - 4a^2 = b^2$

$\displaystyle a^2(b^2 -4) = b^2$

$\displaystyle a^2 = \frac{b^2}{b^2 - 4}$

$\displaystyle a = \frac{b}{\sqrt{b^2 - 4}}$

$\displaystyle A = 4ab$

$\displaystyle A = \frac{4b^2}{\sqrt{b^2 - 4}}$

find $\displaystyle \frac{dA}{db}$ and determine the value of b that minimizes the area.
• Jan 24th 2009, 08:13 AM
Jester
Quote:

Originally Posted by courteous
»Write down the equation of an ellipse, that goes through $\displaystyle P(1,-2)$ and whose circumscribed* rectangle's area is minimized.«
*http://www.shrani.si/f/3e/vc/1Wz3BdGp/ellipse.gif

(Pretending not knowing that "rectangle" is in fact square.)

Set-up:
Ellipse through $\displaystyle (1,-2)$: $\displaystyle \frac{1}{a^2}+\frac{4}{b^2}=1$, rectangle's area: $\displaystyle 2a*2b=min.$
Now, what to differentiate? From the first equation I've also got $\displaystyle a^2b^2=b^2+4a^2$.

You could use Lagrange multipliers. Minimizing $\displaystyle A = 4ab$ subject to $\displaystyle \frac{1}{a^2} + \frac{4}{b^2} = 1$ is to use

$\displaystyle F = 4ab + \lambda \left( \frac{1}{a^2} + \frac{4}{b^2} - 1\right)$

then calculate $\displaystyle F_a,\;\;\;F_b,\;\;\;F_{\lambda}$, set them to zero and solve for $\displaystyle a, b, \; \text{and}\; \lambda$. It avoids the square roots. (You sure about the square?)